ENH: IRR: Add quick exit if all values are of same sign

If all values are of the same sign then no solution exists and
performing Newton's method is redundant. Instead, return `nan`.

Author: Kai-Striega

numpy_financial/_financial.py

@@ -741,6 +741,12 @@ def irr(values, guess=0.1):
     if values.ndim != 1:
         raise ValueError("Cashflows must be a rank-1 array")
 
+    # If all values are of the same sign no solution exists
+    # we don't perform any further calculations and exit early
+    same_sign = np.all(values > 0) if values[0] > 0 else np.all(values < 0)
+    if same_sign:
+        return np.nan
+
     # We aim to solve eirr such that NPV is exactly zero. This can be framed as
     # simply finding the closest root of a polynomial to a given initial guess
     # as follows:

numpy_financial/tests/test_financial.py

@@ -580,9 +580,12 @@ def test_trailing_zeros(self):
             decimal=2,
         )
 
-    def test_numpy_gh_6744(self):
+    @pytest.mark.parametrize('v', [
+        (1, 2, 3),
+        (-1, -2, -3),
+    ])
+    def test_numpy_gh_6744(self, v):
         # Test that if there is no solution then npf.irr returns nan.
-        v = [-1, -2, -3]
         assert numpy.isnan(npf.irr(v))
 
     def test_gh_15(self):